Finding Iterative Solutions to Non-Linear Equations

[topsquark]

I am currently trying to find an iterative solution to the non-linear differential equations represented by
[math]\ddot{ \theta} = A~\cos( \theta )[/math]

and
[math]\dot{ \theta } ^2 = B( \sin( \theta ) - \sin( \theta _0 ) )[/math]

where the dot represents a time derivative and [math]\theta _0[/math] is the angle at time t = 0. (This is the harmonic oscillator where the angle is not taken to be small. A and B are related constants and I can give the derivations if you feel you need them.)

I'm looking for an iterated solution for [math]\theta (t)[/math], but I'm actually more interested in [math]t( \theta )[/math] for now.

Most of the Physics is involved with the first equation, which you are more likely to find as [math]\ddot{ \theta } = A~\sin( \theta )[/math] if you look it up. The second equation can be taken simply to mean that [math]\sin( \theta _0 ) \leq \sin( \theta )[/math] at all times.

Thanks for any help!

-Dan

 

[topsquark]

Just to be clear I am looking for a set of points [math]( t_n, \theta _n )[/math] where [math]\theta _{n+1} = f( \theta _i, t_i )[/math] ([math]0 \leq i \leq n[/math] ) such that the points on the graph are "close" to the function given by the differential equation, so I'm not looking at the elliptic function repesentation of the solution to the differential equation.

-Dan

 

[I like Serena]

Hi Dan,

I'm still not sure what you are asking for exactly.

Anyway, the method to solve an ODE like $\ddot\theta=A \cos\theta$ iteratively is by writing it as $\dot y = f(t,y)$.
In our case:
$$\begin{cases}\dot\theta = \omega \\ \dot\omega = A\cos\theta \end{cases} \Rightarrow \d{}t (\theta,\omega) = f(t,(\theta,\omega)) = (\omega,A\cos\theta)$$
The quick-and-dirty method to solve it is with Euler, which is $y_{n+1}=y_n+h f(t_n, y_n)$:
$$\begin{cases}t_{n+1}=t_n+ h \\ (\theta_{n+1},\omega_{n+1}) = (\theta_{n},\omega_{n}) + h f(t_n,(\theta_n,\omega_n))\end{cases}\Rightarrow\begin{cases}
t_{n+1}=t_n+ h \\
\theta_{n+1} = \theta_{n} + h \omega_n \\
\omega_{n+1} = \omega_{n} + h A\cos\omega_n\end{cases}$$
However, Euler is known to be unstable and yield errors that are ever getting worse.

[box=yellow]Instead, the commonly used method to solve $\dot y = f(t,y)$ is Runge-Kutta:
\begin{cases}
t_{n+1} &= t_n + h \\
y_{n+1} &= y_n + \tfrac{1}{6}\left(k_1 + 2k_2 + 2k_3 + k_4 \right)\\
\end{cases}
where:
\begin{cases}
k_1 &= h\ f(t_n, y_n) \\
k_2 &= h\ f\left(t_n + \frac{h}{2}, y_n + \frac{k_1}{2}\right) \\
k_3 &= h\ f\left(t_n + \frac{h}{2}, y_n + \frac{k_2}{2}\right) \\
k_4 &= h\ f\left(t_n + h, y_n + k_3\right)
\end{cases}[/box]

Alternatively, we can use the same methods to solve $\dot θ^2 = B(\sin(θ)−\sin(θ_0))$.
(We can derive this equation from the previous equation if we assume that $\dot θ_0 = 0$.)
$$\dot θ = f(t,θ) = \pm\sqrt{B(\sin(θ)−\sin(θ_0))}$$
We can apply Euler or Runge-Kutta again as desired.
To instead find $t(\theta)$, we can apply the inverse function theorem, and do:
$$t'(θ) = \frac1{\dot θ} = g(θ, t) = \frac{\pm 1}{\sqrt{B(\sin(θ)−\sin(θ_0))}}\Rightarrow
\begin{cases}
\theta_{n+1}=\theta_n+ h \\
t_{n+1} = t_{n} + h g(\theta_n,t_n)\end{cases}$$
And now apply Euler or Runge-Kutta.

 

[topsquark]

Thank you. I worked with Runge-Kutta once and was able to use it all right, but I really don't understand where it comes from. In fact my otherwise fairly exhaustive personal library doesn't cover this topic at all and I'm wanting to understand what I'm up to. For that reason and one other I've decided to try to do a Taylor expansion recursion formula (that's what I've been meaning when I've used the word "iterative.") By doing this I should be able to match my series with the series for the associated elliptic integral (which I also have few sources about.)

Thanks for the information and advice. I'm at my parents at the moment and I'm not working consistently on it. I'll probably get back to you folks here in a week or so.

-Dan

 

[topsquark]

I gave up on my project.

I was eventually able to derive the Taylor series. In fact I was a bit embarrassed how simple the solution was and how long it took me. Ah well.

Anyway I thought I'd let you all know what I was up to in case anyone wants to play with it. I was in the initial stages of finding an iterative solution to the double pendulum problem. (I was trying out methods on the simple pendulum.) I wanted to see if I could get any information about modes via Fourier transforms. I just derived the "full" equations for the double pendulum, as opposed to the small angle approximation, and the two equations of motion go all the way across the page. Not something I'd want to set up by hand, Runge-Kutta or not. I could program the solution but then I'd have a page full of data to try to Fourier transform, which would have to be done by computer as well. I figure I wouldn't get any inspiration from a page full of numbers, so I've quit.

It was fun while it lasted!

-Dan